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Billionaire Offers $1 Million to Solve Math Problem

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posted on Jun, 7 2013 @ 12:44 PM
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reply to post by odiez2
 


Actually, the article I linked presented the info wrong. It is my understanding that when you post something in Breaking Alternative News, you have to quote the article and its title. That is all I did.




posted on Jun, 7 2013 @ 01:36 PM
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reply to post by TeflonBear
 


dang, some asian kid already won by now....



posted on Jun, 7 2013 @ 01:47 PM
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It all started with Beal's determination to solve the 350-year-old mystery of Fermat's Last Theorem - the idea that Ax + By = Cz. Beal realized that there could only be solutions to the equation when A, B, and C have a common numerical factor.


Are we just supposed to prove the theorem?

Ax + By = Cz
A= 2
B= 6
C= 4
X= 1
Y= 3
Z= 5

2(1) + 6(3) = 4(5)

I even challenged myself to not use any number more than once.

I don't understand if it's supposed to be more complicated than that. lol And if that's it, where's my million?


ETA: Oh, this article doesn't give all the info. There is more to it than that and it also appears to be exponents.


edit on 6/7/2013 by AshleyD because: (no reason given)



posted on Jun, 7 2013 @ 02:05 PM
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I predict this will eventually be solved by some genius prodigy child or teenager, who may not even be aware of the rewards for the correct answer. Someone, eventually, will have a mind wired just right to look at it and say "Ah-HA! That's EASY! Here's the answer, silly!"

...While the rest of us sit back, groan and go back to feeling like complete, hopeless idiots in the face of true brilliance. It's happened before, I believe...and that's how I figure we'll here about this ending. That special someone who just sees the world differently than everyone else.



posted on Jun, 7 2013 @ 02:11 PM
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Ax + By = Cz
z- C + Ax - By
-zC + Ax = By
-zC + AB - xy = 0

Answer is: 0



posted on Jun, 7 2013 @ 02:18 PM
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reply to post by AshleyD
 


These are not multiplicated with each other, but powered.

A powered of x + B powered of y=C powered of z . x,y,z must all be larger than 2

In order to win the prize you need to find such combination, where there is no common prime factor and debunk the conjecture by that

OR

Prove that absolutely in every single such case, where some numbers fit the equation, all the numbers have same prime factor. (Prime numbers=numbers that divide with only themself and one: 2,3,5,7,11,13,17,19,23,29,31 and so on. ).
edit on 7-6-2013 by Cabin because: (no reason given)

edit on 7-6-2013 by Cabin because: (no reason given)



posted on Jun, 7 2013 @ 02:26 PM
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reply to post by Wrabbit2000
 


I doubt it, to be honest. There have already been many geniuses working on it for nearly 2 decades. This was proposed in the first part of 1990s and is quite famous.

For example one of the most famous theorem: Fermat´s last theorem, which is where this one came from:

A^n+B^n=C^n, n>3 Fermat claimed that there are no n-s that satisfy this theorem. This statement was proven in 1993 by Wiles.

It was very famous theorem among mathemathicians and it took 358 years to solve it. There were many prizes for it and thousands of incorrect proofs were submitted before it was finally proved in the 1990s.



posted on Jun, 7 2013 @ 02:38 PM
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this is really thrashing my head now..... heres my attepted work... which i also dont understand lol

A x + B y = C z

1(24) + 2(25) = 3(26)
(Real answer) ~ 349

1 x 24 + 2 x 25 = 3 x 26?



posted on Jun, 7 2013 @ 03:46 PM
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reply to post by Cabin
 


Yeah, I noticed that after reading about it in more thorough articles after thinking it couldn't be that easy. lol

The article cited in this thread doesn't mention the fact exponents are involved and instead makes it appear as multiplication. Then, this article doesn't list the exceptions and restrictions for the theorem. Not a very thorough article.

Oh well, no million for me today.



posted on Jun, 7 2013 @ 04:09 PM
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Andrew Wiles solved this theorem in 1995. Submitted in two manuscripts, "Modular elliptic curves and Fermat's Last Theorem" and "Ring theoretic properties of certain Hecke algebras". So the money should be awarded to Mr. Wiles, I suppose.

Oops, Cabin beat me to it.
edit on 7-6-2013 by eManym because: (no reason given)



posted on Jun, 7 2013 @ 04:10 PM
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I know! it is ("STOP!!!!! delete delete. I think I will get MY Money first!")

a b c and x y z
this is a joke!



posted on Jun, 7 2013 @ 05:42 PM
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Originally posted by Dustytoad
reply to post by TeflonBear
 


Are you guys silly? 3 variables times 3 variables arranged in an equation..

There is no answer..

It's a lot like life, actually..

He'd bet more than 1/8 thousandth of his worth if he expected something.

It's only there to make you think in 3 dimensions, which is the future of rational thought. Not that I can do that coherently either..


edit on 6/7/2013 by Dustytoad because: (no reason given)


Stop rationalizing and just find the answer, maybe something close, there is prize money here. It is an equation made by humans so it can be broken by humans.

Like a lotto ticket though, if you win will someone hunt you down and try to destroy you for knowing? The answer could unlock something that sets off a now global chain of events that leads to something Fermat was afraid to see.

It's is not a three by three. It could be a nine by nine by nine to start thinking about it. It requires math where there is no zero, and you'd probably need all the letters of the alphabet, because numbers plus letters equals crazy. All I'm thinking about is gametes and DNA now, where Ax plus By equals Cz because A and B got married and had an alien baby with a z chromosome.

I swear if you find this and plug it into a machine Fermat pops out of a wormhole.
edit on 7-6-2013 by Sandalphon because: Typos in the equation obviously



posted on Jun, 7 2013 @ 06:12 PM
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The contest must be for the values ABC and xyz that make the equation correct. Just six values, three multipliers with three numbers, when two value combination are added gives the third. Seems deep, very deep. Since its been proven there is a solution.

Of course, each three letter combination has to have three different values.

Ummm....

A=3
B=6
C=4
x=2
y=1
z=3


Yeah....that's it.

edit on 7-6-2013 by eManym because: (no reason given)

edit on 7-6-2013 by eManym because: (no reason given)



posted on Jun, 7 2013 @ 06:12 PM
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This is a job for Sheldon Cooper



edit on 7-6-2013 by NeoSpace because: (no reason given)



posted on Jun, 7 2013 @ 06:26 PM
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According to the website;


If Ax + By = Cz, where A, B, C, x, y and z are positive integers and x, y and z are all greater than 2, then A, B and C must have a common prime factor.


Unique integers and unique exponents.. appears difficult. Let me try and give it a shot..

Ax + By = Cz
A(x) + B(y) = C(z)

A = 100
B = 25
C = 60

While..
X = 3
Y = 12
Z = 10

Plugged in, we get:

100(3) + 25(12) = 60 (10)

100 x 3 is .. 300
25 x 12 is .. 300
60 x 10 is .. 600


Therefore, 300 + 300 = 600

And again, the rules state..

where A, B, C, x, y and z are positive integers

Check, all numbers are positive (and unique) ..

x, y and z are all greater than 2

Check, x = 3, y = 12, and z = 10, all unique and greater than 2 ..

then A, B and C must have a common prime factor

A = 100, B = 25, C = 60..
The common prime factor of all three is 5 ..
[[(100 is [(2 x 2) x (5 x 5) --> (4 x 25) --> 100],
25 is [(5x5)], and
60 is [(2 x 2) x 3 x 5] --> (4 x 3 x 5), while 4 x 3 is 12, so 12 x 5 --> 60)]]

The website states,

The prize is now this: $1,000,000 for either a proof or a counterexample of his conjecture.

As far as I can tell, I provided a proof for this theorem, but not a counter example. Though it does not state I need a counter example..

Am I missing something? That seemed way too easy.

Someone please correct me if I'm wrong.
And God bless, all you critical thinkers.
edit on 7-6-2013 by Lionhearte because: (no reason given)

edit on 7-6-2013 by Lionhearte because: (no reason given)



posted on Jun, 7 2013 @ 06:38 PM
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Wouldn't it involve two independent number sets of infinite lengths as the solution? One set for ABC and the other for xyz such as one set consisting of subset of the prime number set, for every third prime number and the second set, a subset of the prime number set of every fifth prime number. You could also throw in subsets of the Fibonacci number set instead of the prime numbers or maybe a combination of the two.
edit on 7-6-2013 by eManym because: (no reason given)



posted on Jun, 7 2013 @ 06:47 PM
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Originally posted by iwontrun
Come on guys the answer is.......purple!
edit on 6/7/2013 by iwontrun because: (no reason given)


Oooh, we are on similar thoughts, I thought the answer was lilac.........



posted on Jun, 7 2013 @ 07:01 PM
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Originally posted by Wrabbit2000
I predict this will eventually be solved by some genius prodigy child or teenager, who may not even be aware of the rewards for the correct answer. Someone, eventually, will have a mind wired just right to look at it and say "Ah-HA! That's EASY! Here's the answer, silly!"


I'm a genius prodigy (old fart who acts like a) child!! I said the same answer on page one! I told you all the answer was EASY! (cz)



posted on Jun, 7 2013 @ 07:08 PM
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reply to post by TeflonBear
 

he wants a numerical solution?

i really don't understand what he wants....all i see are easy ways to complete it.

(3x4)+(2x6)=(1x24)

or 12+12=24

(6x8)+(4x12)=(3x32)

all unique variables
edit on 7-6-2013 by Bob Sholtz because: (no reason given)



posted on Jun, 7 2013 @ 07:15 PM
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Originally posted by Cabin

If you can find one solution where A,B and C do not have same prime factor you have earned million.



I'd be surprised if they didn't run it through a computer program to come up with all the solutions.

Meaning: Don't waste your time looking for a counter example.



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